Silver chloride, \(\mathrm{AgCl}(s)\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(\mathrm{C}\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$\mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}_{2}^{-}(a q)$$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}^{-}\) is soluble. At \(25^{\circ} \mathrm{C},\) is the solubility of AgCl in a \(0.100 M\) NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\) ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations.

Step by step solution

01

Equation for the dissolution of AgCl in water

The dissolution of silver chloride (AgCl) in water can be represented by the following chemical equation: $$\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q)$$

02

Expression for \(K_{c}\) for the dissolution of AgCl in water

The chemical equation in step 1 represents an equilibrium reaction. We can now write the expression for the equilibrium constant Kc as the product of the concentrations of the products divided by the reactants' concentrations. Since AgCl is a solid, it doesn't appear in the expression as its concentration doesn't vary during the reaction. $$K_{c} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]$$

03

Predict the solubility of AgCl with increasing temperature using Le Châtelier's principle

To predict the solubility's dependence on temperature, we need to evaluate the reaction's enthalpy change (ΔH). In Appendix C, we find that the enthalpy of formation (ΔHf°) of AgCl(s) is -127.1 kJ/mol. Before we can use Le Châtelier's principle, we need to consider the enthalpy of dissolution. We know that the dissolution of AgCl is endothermic, so it absorbs heat in the process (positive ΔHdissolution). Le Châtelier's principle states that a system in equilibrium will counteract any changes in temperature, pressure, or concentration imposed on it. Since the dissolution of AgCl is endothermic, increasing the temperature will favor the dissolution process, and therefore the solubility of AgCl in water will increase.

04

Analyze the effect of a soluble complex in the solubility of AgCl

In the presence of chloride ions in the solution, Ag+ ions may form a soluble complex with Cl- ions as follows: $$\mathrm{Ag}^{+}(a q)+2\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q)$$ The equilibrium constant for this reaction is given as \(K_{c}' = 1.8\,x\, 10^{5}\) at 25°C. In a 0.100 M NaCl solution, the concentration of Cl- ions is 0.100 M. Now, let's write the expression for the equilibrium constant of the given reaction: $$K_{c}' = \frac{[\mathrm{AgCl}_{2}^{-}]}{[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]^{2}}$$ We know the solubility of AgCl in water is \(1.6\, x\, 10^{-10}\) at 25°C. Let's call the solubility 'x': $$K_{c} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] = x(0.100 + x)$$ As x is small compared to 0.100, we can ignore \(x^{2}\), so we have x = \(1.6\, x\, 10^{-10}\). Now, let's find the concentration of AgCl2-: $$K_{c}' = \frac{[\mathrm{AgCl}_{2}^{-}]}{x(0.100)^{2}}$$ $$[\mathrm{AgCl}_{2}^{-}] = K_{c}' * x * (0.100)^{2} = 1.8\, x\, 10^{5} * 1.6\, x\, 10^{-10} * 0.01 = 2.88\, x\, 10^{-7}\, M$$ Since the concentration of the complex AgCl2- (\(2.88\, x\, 10^{-7}\) M) is much greater than the solubility of AgCl in water (\(1.6\, x\, 10^{-10}\) M), it shows that the formation of the soluble complex does, in fact, increase the solubility of AgCl in a 0.100 M NaCl solution compared to pure water.

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